PARTICLE

DYNAMICS OF A
128
2
3

2 1

3 2ga

velocity
projected vertically upwards with a heights
Ex, 7. A particle issurface. If h and H beuniform
the greatest
and variable
Vror. the earth'sparticle moving under
attained by therespectively show
that
accelerations 1 1
h H R [C.H. 67]
earth. greatest height
the radius of the acceleration the
where R is
with uniform
Ans. For motion .

attained is h. 0= V'-2gh ...(1)

y2=2gh the distance
of
where x is the
Or, attraction is we time t. Then
Again, when the earth's centre 0, at any
from the (u > 0).
the particle is i -ue.
equation of motion
surface of the earth
g=/R or, u =gR
On the
i=-gR?
du--gRl2
Or, dx
2gR2
2u du = -dx.
or,

Integrating, we get
2= 2gR'x + A
2gR?
At x = R+H, v =0; . A = R+H

1
R+H)
 MOTION IN ASTRAIGHT LINE 129
Again, when x = R, v= V (the velocity of
partiele from the Earth's projection of the
surface), we get
y²- 2gR' R1 1 ..(2)
R+H)
From (1) and (2), we get
1 1
2gh = 2gR? R R+ H

h= R2.R+ H-R RH
RR+h) R+H
or, 1 1 .1
h R H
1 1 1
or,
h H R
Ex. 8. A unit particle is attracted by two centres of forces
Aand B, each of which attracts it with a force / at distance
x. If the .particle is initially at rest at a point in AB produced,
distance a3 from the middle point of AB, show that it will arrive
at B after a time

[where AB = 2a]

Ans. Choose the middle point of AB as the origin. Let P be

the position of the particle
at any time t, then the A P B D
forces of attraction per Fig. :5.8
unit mass to the centres A
and B are respectively a/(a + *)° towards 0 and /(a-x)³ away
from O.
The equation of motion is

(a-) (x+a)'
du
or, dx (a-x (x+a

DP -9
 DYNAMIs OF APARTICLE

(a-'(r.a
Iniegrating. we get
1 1
|+C
PM-n(a'
Initially x=3a, v= 0.

1 1
0 = +C

0= + C
Or,
(3-1)"
or, 0= +C
a

2
1 1 2

2a{ +a')-2*-2a: +a)

2
(-af
dx
dt Va?
 MOTION IN ASTRAIGHT LINE 131

Integrating between the limits ( = 0 when IN34

x= to l= ,
when r =4, We get

24
Va2 {3a.

Ja(3a' -)
Let x=/3 asin0. Also when x = a, = sin V3
and when
x=W3a, 0=/2.
1
1 3 sin --:sin

/2
t = =[-3cos- log(cosec® -cot)i15

sin
1-cost Jsin-143

-log 1- cos sin 1/J3

1 W3
1-42N3
1 1

Ex. 9. A Particle is let fall from rest from a point outside

the earth at a distance b from the centre. Prove that the square
132 DYNAMICSOF A PARTICLE
of the velocity of the particlo on reaching the centre is
2a

where a is the radius of the earth and g is the value of gravity

at its surface. [C.H. '84]
Ans, Outside the earth's surface acceleration due to gravity
IS inversely proportional to the square of the distance from centre
of the earth. Thus if x be the distance of the particle, then the
equation of motion is
(u > 0) ..(1)

On the earth's surface,

&=/a or, u = a'g
...(2)

Multiply both sides of (2) by 2i and integrating, we have

2 =2ga'lx +c
Initially, £ =0 when x =b.
c=-2ga²b
Hence & = 2a'g (1l/x - 1/6).
If ube the velocity on the earth's surface x = a, then
u= 2ag (1- a/b) ...(3)
Again, inside the earth's surface acceleration at any point is
proportional to its distance from the centre of the earth. Thus
if x be the distance of the particle from the earth's centre then
the equation of motion is
(ñ'>0)
On the surface of the earth
g='a
p'=gla

Hence ...(4)

Multiplying both sides by 2x and integrating, we get

 MOTION IN A STRAIGHT LINE 133

Since at x = a, =U, .. C = U+ ga.

Or, C =2ag(1 - a/b) + ga= ag(3 - 2a/b) [using (3)]
2a
=ag 3
which gives the velocity (inside the earth) at a distance x from
the centre of the earth.
Hence the velocity vat the earth's centre ( = 0) is
2a

Ex. 10. Aparticle starts from rest at a distance b(> a) from

a fixed point O, under the action of a force through the fixed
point, the law of which at a distance x from O is u(1 - al) towards
0 when xis greater than a, but it is u(ax"- ax) away from
Owhen x is less than a. Show that the velocity of the particle
will again vanish when x = ab. [C.H. '01]
Ans. Let OA =a and
OB = b. At any time t, let A P

Pbe the position of the Fig.: 5.9

particle such that OP = x.
Then the equation of motion is
* = -(1 - alx) [for x > a) ..(1)

Multiplying both sides by 2* and integrating, we get

2=-2u(x - alogx) + c
Initially, x=b, i =0; .:. c=2u(b-a log b).
i = 2u(b-x) - 2ua log (b/)
Let i =u when x = a,
u'= 2u(b-a) - 2ua(bla) ...(2)
Again, when x <a, then the equation of motion is
...(3)

Multiplying both sides by 2* and integrating, we get

i =-2u(a'lx + alog x) + c ..(4)
At x = a, =v given by (2)
u=- 2åa - 2ualoga + c'
c= 2u(b - a)- 2ualog bla + 2ua + 2ualog a
= 2ub - 2uclog (b/a')
 134 DYNAMICS OF A PARTICLE

Substituting this value of c' in (4), we get

'=- 2uíak - b) + 2ualog (a'b) - 2ualog x
or, --a?lb)+2ualog bx
2
|2ub| a
-2ualog bx
which shows that the velocity of the particle again vanishes when
x= a'b.
Ex. 11. A particle moves in a straight line from a distance
a to a centre of force. Show that the time required to reach the
centre is atNu or 2a3u according as the force of attraction
varies inversely as the cube of the distance or as the 5/3 th power
of the distance, being the constant of variation.
Ans. First consider the motion when the force of attraction
varise inversely as the cube of the distance x from the centre
of force 0. The equation of motion is

Multiplying both sides by 2* and integrating, we have

Initially, | =0 at x= 0; .:. c=-la.

1 1

va'-2
Or, ..(1)

[negative sign is taken, since the particle is moving towards the

origin.]
Integration of (1) gives, the time t, when x=0 as
axdx
dt =
VAVa-?
 MOTION IN ASTRAIGHT LINE 135

When the force of attraction varies inversely as the 5/3 th

power of the distance, the equation of motion is

5/3

Multiplying both sides by 2* and integrating, we get

3u +C

3_u
Initially, x = 0 at x = a, .: c = 2/3

1 1
=34 a2/3

or, Va8,1/3
al3./3
1
dt =
or, 2/3

Integrating between the limits t =0 when = ato t=t, when

x =0, we get

,2/3 2/3

Substituting x = acos9, we have

3a4/3
cos 9d
t, = Jo

2a43
-sin°alosbd9
Hence the result.
a fixed point varying
Ex. 12. A particle is attacted by a force to the velocity acquired
distance; if
inversely as the nth power of thedistance to a distance 'a' from the
by it in falling from an infinite acquire in falling from rest at a
centre be equal to that it would that n = 3/2. [VH. 01)
distance 'a' toa distance 'al4', then show
 136 DYNAMICS OF APARTICLE

Ans. The equation of motion is

Multiplying both sides by 2* and integrating, we get

-n+1
+C
(n-1) ...(1)

When x’ 0, =0 .. c=0
2u
(r-1)*-1 ..(2)
Let v, be the velocity when the particle reaches at A.

(n-1)a-! ..(3)

Again, when x = a, = 0 (when particle falling from a)

0= 2u
(n-1)a
Or, C= 2u
(n-1)a*-1
Hence (1) becomes
1 1
n-1 -1
Let be the
B (n = a/4) velocity when the particle reaches at
1
n-1| a-1 ...(4)
From the given condition, we have
Or, 2 D2
2u 2u
(n-)a4-1 (2-1)a-+|4"-1)
or,
4n-l-1=1
 MOTION IN ASTRAIGHT LINE 137

or, 4n- = 2
or, g2n-2 = 2
or, 2n - 2 = 1
n= 3/2
Hence the result.

" Exercise V

A particle initially at rest, moves form a fixed poiont in a straight

line so that at the end of t secs., its acceleration is
1
sint +
(t+1)"
Show that its distance from the fixed point at the end of n secs is
2 log(1 + 1)
fixed point 0 with
2. A particle moves in a straight line towards a
from rest at a distance a
an attractive force u x (distance), If it starts distance a/2 form 0 is
from O, Show that the time of reaching a
-/3u and that its velocity is then
2
straight line, is acted on by a
&.AParticle of mass m moving in aIf it starts from rest at a distance
origin 0.
force mu(x t a / ) towards the
when x = a/2 is given by
afrom 0, show that its speed
)/2
15
4

from infinity; show that its

towards the earth
4. A particle falls surface of the earth is the same as that which
velocity on reaching the falling with constant acceleration g through
it would have
acquired in
the earth's radius.
a distance equal to fixed point 0
velocity uft/sec from aacceleration
projected with ()
5. A particle is and moves along the line so that its leaves O and
on a straight line time from the instant it
where t is the of acceleration.
is given by kvt, Express the distance from 0in terms the acceleration
k is a constant. value of k be 2, find uin order thatthrough 1ft.
If the numerical 1 ft/sec, when it has moved
shall be
of the particle 4f3 I5 59
Ans.
fisec
15k
is
and its acceleration at any time t
particle starts from
rest
that its maximum velocity u
prove
are constants;
f-kt wher f and k before it acquires this
described by it
is and that the space
velocity is 15u/16f